Arithmetic Sequence
If the numbers in a sequence follows arithmetic addition or subtraction, we call it arithmetic sequence. For example, 1, 3, 5, 7, \ldots ,33 is an arithmetic sequence.
Its first number, a = 1
Interval, d = 2 [ (second term -first term) or (third term – second term) and so on]
Last number, n = 33
The number of elements in this sequence is not equal to 33 since there are some missing numbers. We can calculate the number of elements in this series by using the following equation:
n^{th} \text{element} = a + (n-1)d \\or, 33 = 1 + (n-1)2 \\
or, 33 = 1 + 2n - 2 \\
or, n = 17
Now, we will calculate the sum of all elements in the above series
1 + 3 + 5 + \ldots + 33 \\Sum, S = \frac{n}{2}\left\{2a + (n-1)d\right\} \\
= \frac{17}{2}\left\{2\times1 + (17-1)\times2 \right\} \\ \\
= 17\times17 \\ \\
= 289Interval can be negative number. For example, for the following sequence calculate number of elements and their summation
23 + 20 + 17 + \ldots + 5 \\ \\
n^{th} \: \text{element} = a + (n-1)d \\or, 5 = 23 + (n-1)(-3) \\
or, 5 = 23 - 3n + 3 \\
or, 3n = 21 \\
or, n = 7
Sum, S = \frac{n}{2} \left\{2a + (n-1)d \right\} \\
= \frac{7}{2} \left\{2\times23 + (7-1)\times(-3)\right\} \\ \\
= 7\times14 \\ \\
= 98Geometric Sequence:
If the numbers in a sequence follow some multiplication (greater or less than 1) factor, then it is known as geometric sequence. For example, 1, 2, 4, 8, \ldots , 256
Here first number, a = 1
Ratio, r = 2/1 = 4/2 = 2 (ratio of second term to first term or third term to second term)
n^{th} \text{term} = ar^{n-1} \\or, 256 = 1\times2^{n-1} \\
or, 2^{n-1} = 2^8 \\
or, n-1 = 8 \\
or, n = 9
Sum, S = a \frac{1-r^n}{1-r} \\
= 1 \frac{1-2^9}{1-2} \\
= 511
Calculate the sum of first 10 elements of the following geometric series
1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots , \ldotsHere, a= 1, r= 1/2
Sum, S = a \frac{1-r^n}{1-r} \\
= 1. \frac{1- \frac{1}{2^{10}}}{1-\frac{1}{2}} \\
= 1.998
If instead of 10 elements, we want to determine the summation of the infinite sequence, then the above equation reduces to
Sum, S = a. \frac{1}{1-r} since \frac{1}{2^n} \text{becomes zero as} n \Rightarrow \infty
= 1. \frac{1}{1-\frac{1}{2}} \\
= 2